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Standard matrix for reflection across the line y=-x

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Let T : R2→R2, be the matrix operator for reflection across the line L : y = -x

    a. Find the standard matrix [T] by finding T(e1) and T(e2)

    b. Find a non-zero vector x such that T(x) = x

    c. Find a vector in the domain of T for which T(x,y) = (-3,5)

    2. Relevant equations


    3. The attempt at a solution

    a. I found [T] =
    0 -1
    -1 0

    b. I'm not really sure what this is asking. Do I just pick a random x = (x1,x2)
    and then plug in T(x) = x?

    c. T(-3, 5) would be (-5, 3). Is that what the question is asking? Is (-5, 3) in the domain of T?

    Help me, I'm stupid :D
     
  2. jcsd
  3. Apr 6, 2015 #2
    a. I got ##\Big( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \Big)##

    b. you want to find a vector that the matrix doesn't move. So you don't pick a random one, you start with T(x, y) = (x, y) & then find a vector that satisfies that equation.

    c. that's right
     
  4. Apr 6, 2015 #3

    LCKurtz

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    I agree.

    If you think geometrically about the reflection you are given, isn't it pretty clear which points wouldn't move?

    Yes.
     
  5. Apr 6, 2015 #4
    Thanks guys!
    How did you get that matrix?

    After fiddling with some numbers to try to get it to work I got T(-1, 1) = (-1, 1) which would be a point on the line y = -x so I guess all the points on the line wouldn't move obviously right?
     
  6. Apr 6, 2015 #5

    LCKurtz

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    Fiddling?? Do you mean guessing? Experimenting?? What happens to any point on the line ##y=-x## when you reflect it in that line? Or if you solve$$
    \begin{bmatrix}
    0& -1\\
    -1 & 0
    \end{bmatrix}
    \begin{bmatrix} a\\b\end{bmatrix} =
    \begin{bmatrix} a\\b\end{bmatrix}
    $$for ##a## and ##b##? No guesswork needed.
     
    Last edited: Apr 7, 2015
  7. Apr 7, 2015 #6
    It reflects onto the line but in the opposite quadrant?
     
  8. Apr 7, 2015 #7

    LCKurtz

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    No. Think of your line as a mirror. Any point some distance away on one side of the mirror reflects to appear the same distance away on the other side of the mirror. What about a paint dot right on the mirror. Where does its reflection appear to be?
     
  9. Apr 7, 2015 #8
    It doesn't move then right? Its reflection is on its original location?
     
  10. Apr 7, 2015 #9
    Never mind I did the matrix for the wrong transformation o0) it made sense at the time anyway
     
  11. Apr 7, 2015 #10

    HallsofIvy

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    Here's how I would do that problem: Any 2 by 2 matrix can be written as [tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}[/tex]. "Reflecting about the line y= -x" the vector <1, 0> is mapped into < 0, -1> and the vector <0, 1> is mapped into <-1, 0>.

    So we must have [tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix} 0 \\ -1\end{bmatrix}[/tex] and [tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix} -1 \\ 0\end{bmatrix}[/tex].

    Doing the matrix multiplications on the left and setting the components equal to the right gives you four equations to solve for a, b, c, and d.
     
  12. Apr 7, 2015 #11

    LCKurtz

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    Right. So for that reflection, it's pretty obvious geometrically which points in ##R^2## give ##T(x) = x##.
     
  13. Apr 7, 2015 #12
    Cool thanks, that makes sense! I learned everything I know about math from U of A, its not my fault :D
     
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